Examples of torsion-free abelian groups with finite automorphism group $\mathbb{Z}$ is a torsion-free abelian group with finite automorphism group. Are there other examples of such groups? 
Jumping from $\mathbb{Z}$ to $\mathbb{Q}$ is not good; since $\mathbb{Q}$ has infinite automorphism group. Is there any group in-between $\mathbb{Z}$ and $\mathbb{Q}$ with required property?
 A: If $G$ is a finitely generated abelian group, then $G \cong \mathbb{Z}^n\oplus\mathbb{Z}_{p_1^{n_1}}\oplus\dots\mathbb{Z}_{p_k^{n_k}}$ for some (not necessarily distinct) primes $p_1, \dots, p_n$, and non-negative integers $n, n_1, \dots, n_k$. If $G$ is also torsion-free, then we have $G \cong \mathbb{Z}^n$ and therefore $\operatorname{Aut}(G) \cong GL(n, \mathbb{Z})$. Note that $GL(1, \mathbb{Z}) = \{-1, 1\}$ which is finite, but $GL(n, \mathbb{Z})$ is infinite for $n \geq 2$. So the only finitely generated, torsion-free abelian group with finite automorphism group is $\mathbb{Z}$. 
I'm not sure about the non-finitely generated case.
A: Exercise: If $f$ is any map from the set of primes to the set of non-negative integers, then (1) the subgroup $B_f$ of $\mathbf{Q}$ generated by $1/p^{f(p)}$ when $p$ ranges over primes has automorphism group reduced to $\{\pm 1\}$.
And (2) $B_f$ and $B_g$ are isomorphic if and only if $f,g$ coincide outside a finite subset. (In particular these are uncountably many non-isomorphic groups.) 
